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Q.

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates with the same constant rate and comes to rest. If the total time taken is 4 sec, the distance travelled is

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a

16 m

b

None of the above

c

4m

d

32 m

answer is B.

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Detailed Solution

$\begin{array}{l} 8 = {at}_{1} and 0 = 8 - a (4 - t_{1}) \\ or t_{1} = \frac{8}{a} ∴ 8 = a (4 - \frac{8}{a}) \end{array}$
8 =4 a-8 or a=4
and t₁ = 8/4 = 2sec
Now, $s_{1} = 0 \times 2 + \frac{1}{2} \times 4 (2)^{2} or s_{1} = 8 m$
$s_{2} = 8 \times 2 - \frac{1}{2} \times 4 \times (2)^{2} or s_{2} = 8 m$
$∴ s_{1} + s_{2} = 16 m$

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